Ask question

Ask question

All categories

Svetllana [295]

4 years ago

8

a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an

d school

1 answer:

Rom4ik [11]4 years ago

4 0

<h2>Answer:</h2>

The distance will be 1800 m

<h2>Explanation</h2>

As in question

Time = 15 min

Time = 15 x 60 sec = 900 sec

Speed = 2 m/s

We know that

$Distance = Speed x Time$

$Distance = 2 x 900$

$Distance = 1800 m$

So, the answer is 1800 m

You might be interested in

The rebirth of science and learning during the fifteenth century is termed the

Pepsi [2]

The Renaissance, was the rebirth to science as well as many other advancements

5 0

4 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago

A device that converts mechanical energy into electricity is a

IgorLugansk [536]

Answer:

An electric generator is a device that converts a form of energy into electricity.

Explanation:

7 0

3 years ago

A<br> B <br> C <br> D<br><br> Plz help me.

Montano1993 [528]

Answer:

The correct option is;

${}$ Man doing most work ${}$ Box gaining the most energy

D ${}$ Y Q

Explanation:

The given parameters of the question are;

The distance the box P is pushed by the Man X = 0

The force the Man X applies to the box P = x N

The distance the box Q is lifted by the Man Y = h > 0 meters

The minimum force the Man Y applies to the box Q = W, the weight of the box

Work done = Force × Distance

Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²

The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)

Therefore, Kinetic energy = 0 J

The work done by Man X = 0 × x = 0 J

The energy gained by the box P = W × 0 = 0 J

The work done by Man Y = W × h = W·h J

The energy gained by the box P = W × h = W·h J

We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J

The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J

Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.

8 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago